Question Number 149596 by iloveisrael last updated on 06/Aug/21

Without L′Hopital lim_(x→π/7) ((sin x sin 2x sin 3x−((√7)/8))/(x−(π/7))) =?

Answered by EDWIN88 last updated on 07/Aug/21

let x−(π/7)=t ; x=t+(π/7) ∧ t→0 lim_(t→0) ((sin (t+(π/7))sin (2t+((2π)/7))sin (3t+((3π)/7))−((√7)/8))/t) =lim_(t→0) ((8sin (t+(π/7))sin (2t+((2π)/7))sin (3t+((3π)/7))−(√7))/(8t)) =lim_(t→0) ((−4sin (2t+((2π)/7)){cos (2t+((2π)/7))−cos (t+(π/7))}−(√7))/(8t)) =lim_(t→0) ((−2sin (4t+((4π)/7))+4sin (2t+((2π)/7))cos (t+(π/7))−(√7))/(8t)) =lim_(t→0) ((−2sin (4t+((4π)/7))+2(sin (3t+((3π)/7))+sin (t+(π/7)))−(√7))/(8t)) =lim_(t→0) ((−2sin (4t+((4π)/7))+2sin (3t+((3π)/7))+2sin (t+(π/7))−(√7))/(8t)) =lim_(t→0) ((−8cos (4t+((4π)/7))+6cos (3t+((3π)/7))+2cos (t+(π/7)))/8) =−8cos ((4π)/7)+6cos ((3π)/7)+2cos (π/7) (•)cos ((4π)/7)=cos (π−((3π)/( 7)))=−cos ((3π)/7) =8cos ((3π)/7)+6cos ((3π)/7)+2cos (π/7) =14cos ((3π)/7)+2cos (π/7) L=((14cos ((3π)/7)+2cos (π/7))/8) ≈ 0.614654